LeetCode-116. Populating Next Right Pointers in Each Node

Given a binary tree

struct TreeLinkNode {
  TreeLinkNode *left;
  TreeLinkNode *right;
  TreeLinkNode *next;
}

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.

Initially, all next pointers are set to NULL.

Note:

• You may only use constant extra space.
• Recursive approach is fine, implicit stack space does not count as extra space for this problem.
• You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

Example:

Given the following perfect binary tree,

     1
   /  \
  2    3
 / \  / \
4  5  6  7

After calling your function, the tree should look like:

     1 -> NULL
   /  \
  2 -> 3 -> NULL
 / \  / \
4->5->6->7 -> NULL

这题说只能使用额外的常数空间,然后允许递归解决问题,好的那就递归吧

solution:

public class Solution {
    public void connect(TreeLinkNode root) {
        if(root!=null){
           if(root.left==null)//判断当前根节点是否有子节点
               return;
            root.left.next = root.right;
            if(root.next!=null){
                root.right.next = root.next.left;
            }
            connect(root.left);
            connect(root.right);
        }
    }
}

参考样例图可以发现规律,对于同一棵树,左节点的next指向右节点,如果是右节点的话,判断根节点的next是否指向一个节点,

该节点应该为另一个树的根节点,如果是的话,让右节点的next指向该节点的左节点,然后递归对每一个根节点做此处理就可以得到结果.

能做这两项操作的原因是因为题目里说了每一个节点的next默认为null,所以很好判断其是否存在,其次说了该树是一个完美二叉树

即所有叶子节点都在同一层，每个父节点都有两个子节点(注:但是你还是要判断一个每个根节点是否有子节点,有卡ac的测试数据)